[proofplan] We prove the equivalence by translating between the preimage definition of continuity and the neighbourhood definition of continuity at a point. In the forward direction, the required neighbourhood of $x_0$ is the open preimage $f^{-1}(V)$. In the reverse direction, pointwise continuity supplies, for each point of $f^{-1}(V)$, an open neighbourhood contained in $f^{-1}(V)$; the union of these neighbourhoods is exactly the preimage, hence open. [/proofplan] [step:Use the open preimage condition to obtain pointwise neighbourhoods] Assume $f$ is continuous in the sense that $f^{-1}(V) \in \tau_X$ for every $V \in \tau_Y$. Let $x_0 \in X$, and let $V \in \tau_Y$ satisfy $f(x_0) \in V$. Define the set $U \subset X$ by $U := f^{-1}(V)$. By continuity of $f$, $U \in \tau_X$. Since $f(x_0) \in V$, we have $x_0 \in f^{-1}(V)=U$. Also, by the definition of preimage, $f(U) \subset V$. Therefore $f$ is continuous at $x_0$. Since $x_0 \in X$ was arbitrary, $f$ is continuous at every point of $X$. [guided] Assume $f$ is continuous by the open-set definition: for every open set $V \in \tau_Y$, its preimage $f^{-1}(V)$ belongs to $\tau_X$. We must prove the pointwise neighbourhood condition at an arbitrary point. Let $x_0 \in X$ be fixed, and let $V \in \tau_Y$ be an open neighbourhood of $f(x_0)$, meaning $f(x_0) \in V$. The natural candidate for the required neighbourhood of $x_0$ is the full preimage of $V$. Define $U \subset X$ by $U := f^{-1}(V)$. Since $V \in \tau_Y$ and $f$ is continuous, the set $U$ belongs to $\tau_X$. Since $f(x_0) \in V$, the definition of preimage gives $x_0 \in f^{-1}(V)$, so $x_0 \in U$. It remains to check that this neighbourhood maps into $V$. If $x \in U$, then $x \in f^{-1}(V)$, so $f(x) \in V$. Hence $f(U) \subset V$. Thus for this arbitrary $x_0 \in X$ and this arbitrary open neighbourhood $V$ of $f(x_0)$, we have found an open set $U \in \tau_X$ with $x_0 \in U$ and $f(U) \subset V$. Therefore $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous at every point of $X$. [/guided] [/step] [step:Use pointwise neighbourhoods to show every open preimage is open] Assume $f$ is continuous at every point $x_0 \in X$. Let $V \in \tau_Y$ be arbitrary, and define $A \subset X$ by $A := f^{-1}(V)$. For each $x \in A$, we have $f(x) \in V$, so pointwise continuity at $x$ gives a set $U_x \in \tau_X$ such that $x \in U_x$ and $f(U_x) \subset V$. The inclusion $f(U_x) \subset V$ implies $U_x \subset f^{-1}(V)=A$. We claim that \begin{align*} A=\bigcup_{x \in A} U_x. \end{align*} Indeed, if $x \in A$, then $x \in U_x$, so $A \subset \bigcup_{x \in A} U_x$. Conversely, if $z \in \bigcup_{x \in A} U_x$, then $z \in U_x$ for some $x \in A$, and $U_x \subset A$, so $z \in A$. Hence equality holds. Since each $U_x \in \tau_X$ and a topology is closed under arbitrary unions, $A \in \tau_X$. Thus $f^{-1}(V) \in \tau_X$ for every $V \in \tau_Y$, so $f$ is continuous. [guided] Assume now that $f$ is continuous at every point of $X$ in the neighbourhood sense. We must prove the open-set definition of continuity: for every open set $V \in \tau_Y$, the preimage $f^{-1}(V)$ is open in $X$. Let $V \in \tau_Y$ be arbitrary, and define $A \subset X$ by $A := f^{-1}(V)$. To prove $A \in \tau_X$, we show that every point of $A$ has an open neighbourhood contained in $A$, and then express $A$ as a union of such neighbourhoods. Fix $x \in A$. By the definition of $A$, this means $f(x) \in V$. Since $f$ is continuous at the point $x$ and $V$ is an open neighbourhood of $f(x)$ in $Y$, there exists a set $U_x \in \tau_X$ such that $x \in U_x$ and $f(U_x) \subset V$. The condition $f(U_x) \subset V$ is exactly the statement that every point of $U_x$ lies in the preimage of $V$, so $U_x \subset f^{-1}(V)=A$. Now compare $A$ with the union of all these neighbourhoods: \begin{align*} A=\bigcup_{x \in A} U_x. \end{align*} The inclusion $A \subset \bigcup_{x \in A} U_x$ holds because each $x \in A$ belongs to its own neighbourhood $U_x$. The reverse inclusion holds because every $U_x$ was chosen to satisfy $U_x \subset A$. Therefore the displayed equality is valid. Each $U_x$ belongs to $\tau_X$, and the definition of a topology says that arbitrary unions of members of $\tau_X$ again belong to $\tau_X$. Hence $A \in \tau_X$. Since $A=f^{-1}(V)$ and $V \in \tau_Y$ was arbitrary, $f^{-1}(V) \in \tau_X$ for every open set $V$ in $Y$. Therefore $f$ is continuous. [/guided] [/step] [step:Combine the two implications] The first step proves that continuity of $f$ implies continuity at every point of $X$. The second step proves that continuity at every point of $X$ implies continuity of $f$. Therefore $f$ is continuous if and only if $f$ is continuous at every point $x_0 \in X$. [/step]